13 divides 2⁷⁰ + 3⁷⁰

Powers of $2$. $70=5\cdot12+10$, so $2^{70}\equiv2^{10}=1024\pmod{13}$. Since $1024=13\cdot78+10$, $2^{70}\equiv10$.

Powers of $3$. $3^3=27\equiv1\pmod{13}$, and $70=3\cdot23+1$, so $3^{70}\equiv3^1=3$.

Sum. $2^{70}+3^{70}\equiv10+3=13\equiv0\pmod{13}$.

Remark. Another approach: $2^{70}+3^{70}$ is divisible by $13$ because $3^2\equiv-4=-2^2\pmod{13}$, hence $3^{70}=(3^2)^{35}\equiv(-4)^{35}=-2^{70}$.

Lesson. Reduce the base and the exponent: use Fermat's theorem (or the actual order) to shrink exponents modulo $p-1$ (or modulo the order).